2018 AMC 8 Problems/Problem 19

Problem 19

In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?

$[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]$

$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

Solution

Instead of + and -, let us use 1 and 0, respectively. If we let $a$ , $b$ , $c$ , and $d$ be the values of the four cells on the bottom row, then the three cells on the next row are equal to $a+b$ , $b+c$ , and $c+d$ taken modulo 2 (this is exactly the same as finding $a \text{ XOR } b$ , and so on). The two cells on the next row are $a+2b+c$ and $b+2c+d$ taken modulo 2, and lastly, the cell on the top row gets $a+3b+3c+d \pmod{2}$ .

Thus, we are looking for the number of assignments of 0's and 1's for $a$ , $b$ , $c$ , $d$ such that $a+3b+3c+d \equiv 1 \pmod{2}$ , or in other words, is odd. As $3 \equiv 1 \pmod{2}$ , this is the same as finding the number of assignments such that $a+b+c+d \equiv 1 \pmod{2}$ . Notice that, no matter what $a$ , $b$ , and $c$ are, this uniquely determines $d$ . There are $2^3 = 8$ ways to assign 0's and 1's arbitrarily to $a$ , $b$ , and $c$ , so the answer is $\boxed{\textbf{(C) } 8}$ .

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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2018 AMC 8 Problems/Problem 19

Problem 19

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